Answer to Question #334327 in Statistics and Probability for jul

Question

From a finite population consisting of digits 1, 4, 5, and 9, construct the

sampling distribution of the sample mean

a. With a sample size of 2

b. With a sample size of 3

Answer 1

a. We have population values 1,4,5,9, population size N=4 and sample size n=2.

Mean of population $(\mu)$ = $\dfrac{1+4+5+9}{4}=4.75$

Variance of population 

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}$$

$$=\dfrac{14.0625+0.5625+0.0625+18.0625}{4}=8.1875$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{8.1875}\approx2.8614$$

The number of possible samples which can be drawn without replacement is $^{N}C_n=^{4}C_2=6.$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 1,4 & 2.5\\ \hdashline 2 & 1,5 & 3 \\ \hdashline 3 & 1,9 & 5 \\ \hdashline 4 & 4,5 & 4.5 \\ \hdashline 5 & 4,9 & 6.5 \\ \hdashline 6 & 5,9 & 7 \\ \hdashline \end{array}$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) & \bar{X}^2f(\bar{X}) \\ \hline 2.5 & 1/6 & 2.5/6 & 6.25/6 \\ \hdashline 3 & 1/6 & 3/6 & 9/6 \\ \hdashline 4.5 & 1/6 & 4.5/6 & 20.25/6 \\ \hdashline 5 & 1/6 & 5/6 & 25/6 \\ \hdashline 6.5 & 1/6 & 6.5/6 & 42.25/6 \\ \hdashline 7 & 1/6 & 7/6 & 49/6 \\ \hdashline \end{array}$$

Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=4.75=\mu$$

The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{151.75}{6}-(4.75)^2=\dfrac{16.375}{6}= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$$ $$\sigma_{\bar{X}}=\sqrt{\dfrac{131}{48}}\approx1.6520$$

b. We have population values 1,4,5,9, population size N=4 and sample size n=3.

Mean of population $(\mu)$ = $\dfrac{1+4+5+9}{4}=4.75$

Variance of population 

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}$$

$$=\dfrac{14.0625+0.5625+0.0625+18.0625}{4}=8.1875$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{8.1875}\approx2.8614$$

The number of possible samples which can be drawn without replacement is $^{N}C_n=^{4}C_3=4.$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 1,4,5 & 10/3\\ \hdashline 2 & 1,4,9 & 14/3 \\ \hdashline 3 & 1,5,9 & 15/3 \\ \hdashline 4 & 4,5,9 & 18/3 \\ \hdashline \end{array}$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) & \bar{X}^2f(\bar{X}) \\ \hline 10/3 & 1/4 & 10/12 & 100/36 \\ \hdashline 14/3 & 1/4 & 14/12 & 196/36 \\ \hdashline 15/3 & 1/4 & 15/12 & 225/36 \\ \hdashline 18/3 & 1/4 & 18/12 & 324/36 \\ \hdashline \end{array}$$

Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=4.75=\mu$$

The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{845}{36}-(4.75)^2=\dfrac{32.75}{36}= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$$ $$\sigma_{\bar{X}}=\sqrt{\dfrac{32.75}{36}}\approx0.9538$$